What is a universal property?

The following three constructions have something in common:

  • Kernels: If f:G\to H and g:H\to K are two group homomorphisms, then the composite g\circ f is the trivial homomorphism G\to K if and only if the image of f is contained in the kernel of g.
  • Polynomial rings: If A is any \mathbb{R}-algebra, then an \mathbb{R}-algebra homomorphism \mathbb{R}[x]\to A is entirely determined by where it sends x.
  • Topological products: The product topology on a product of topological spaces \prod_{n=1}^\infty X_n is a little counterintuitive—at first glance, the box topology seems more natural—but the product topology has the property that a function (f_1,f_2,\dots):A\to\prod_{n=1}^\infty X_n is continuous if and only if each component map f_n:A\to X_n is continuous.

Do you already see the similarity? Let me rephrase them to make it more obvious:

  • Kernels: If g:B\to C is a group homomorphism, and A is any group, then group homomorphisms A\to \mathrm{ker}(g) are in canonical one-to-one correspondence with homomorphisms A\to B whose composite with g is trivial.
  • Polynomial rings: If A is any \mathbb{R}-algebra, then \mathbb{R}-algebra homomorphisms \mathbb{R}[x]\to A are in canonical one-to-one correspondence with elements of A.
  • Topological products: If X_1,X_2,\dots and A are any topological spaces, then continuous functions A\to \prod_{n=1}^\infty X_n (with the product topology) are in canonical one-to-one correspondence with tuples of continuous functions A\to X_1, A\to X_2, etc.

These are universal properties, and they all have the following rough format:

Given some sort of data one could attach to an arbitrary object A

(e.g. group homomorphisms A\to B whose composite with g is zero, or elements of A‘s underlying set, or a tuple of continuous functions A\to X_n,)

…this type of data comes in one-to-one correspondence with maps from A to some “universal object,” or the other way around.

(e.g. group homomorphisms from A to the \mathrm{ker}(g), or algebra homomorphisms from \mathbb{R}[x] to A, or continuous functions from A to the product space \prod_n X_n.)

In principle, that’s all there is to an object with a universal property: it’s some object such that maps to it from (or from it to) some arbitrary object A correspond to some kind of data you could attach to A.

But the reason this notion is so powerful is that having a universal object for some kind of data—let’s say that the set of all possible data on A is denoted F(A), and that the universal object is X—tells us three related things about F and X:

  1. F(A) varies “functorially” with A. In other words, if we have a map A\to A', we should either consistently get a function F(A)\to F(A') or one F(A')\to F(A).

In the kernels example, we have the “data” function F given by

F(A) = \{\text{group homomorphisms }f:A\to B| g\circ f:A\to C\text{ is trivial}\}.

Given a group homomorphism A\to A', and a piece of data f:A'\to B on A', we get a composite homomorphism A\to A'\to B, whose composite with g is trivial if g\circ f is. So we get a function F(A')\to F(A), a way to “pull back” data on A' to data on A.

We also get a notion of pullback in the case of topological products, for which the data on a space A is a tuple of maps out of A. If (A'\to X_n)_{n=1}^\infty is a tuple of continuous functions, and A\to A' is any continuous map, then the tuple of composites (A\to A'\to X_n)_{n=1}^\infty is again a tuple of continuous functions, so we again can “pull back” data from A' to A.

In both of these cases, from a map A\to A' we get a map F(A')\to F(A) in the “other direction,” but in the case of polynomial rings, the relevant notion of “data” on an algebra A is just an element of A, so F(A) = A as sets. Then from an algebra homomorphism A\to A', we just forget the ring structure and get a function A\to A', letting us “push forward” elements of A to elements of A'. When data can be “pushed forward” along maps A\to A', we say that the data function F is a covariant functor; if such a map gives a “pullback” F(A')\to F(A) instead, we say that the functor F is contravariant.

Question: The two contravariant data functors we’ve considered are the ones for which the maps go to the universal object (the kernel and product), and the covariant data functor is the one where the maps go from the universal object (the polynomial ring). Does it always work out like that?

Answer: Yes! If maps going from A to the universal object X correspond to pieces of data on A, like maps to the kernel or maps to the product, then we have a collection of bijections F(A)\cong \mathrm{Hom}(A,X), the set of maps from A to X. This latter set varies contravariantly in A: if we have a map A\to A', then we get a function \mathrm{Hom}(A',X)\to\mathrm{Hom}(A,X) sending a map A'\to X to the composite A\to A\to X. This gives F a notion of pullback, making it a contravariant functor even if it wasn’t one.

In a similar way, if we have a collection of bijections F(A)\cong\mathrm{Hom}(X,A), as in the case of elements of an \mathbb{R}-algebra A corresponding to \mathbb{R}-algebra homomorphisms \mathbb{R}[x]\to A, then the fact that \mathrm{Hom}(X,A) varies covariantly with A makes F a covariant functor.

So if you have a notion of data you can attach to an object, and are wondering whether it might be parameterized by maps to or from some universal object, first see whether the data varies contravariantly or covariantly.

  1. The universal object X comes with a canonical piece of data x_0\in F(X).

Continuous functions from a space A to the topological product \prod_{i=1}^n X_n correspond to tuples of maps A\to X_1, A\to X_2, etc., and \prod_{i=1}^n X_n comes with such a tuple: the projections \prod_{i=1}^n X_n\to X_1, \prod_{i=1}^n X_n\to X_2, etc.

Given a group homomorphism g:B\to C, group homomorphisms to \mathrm{ker}(g) parameterize maps to B whose composite with g is trivial. And indeed, \mathrm{ker}(g) itself has a canonical group homomorphism to B, the inclusion of it as a subgroup, and the composition of that homomorphism with g is trivial.

And for the case where the data on an \mathbb{R}-algebra A is an element of A, the universal object \mathbb{R}[x] has a distinguished element: x.

In general, if F is the notion of data and X is the universal object, we get a distinguished element x_0 of F(X). This always happens because, whether the data F comes with bijections F(A)\cong \mathrm{Hom}(A,X) or F(A)\cong \mathrm{Hom}(X,A), we always get a bijection F(X)\cong \mathrm{Hom}(X,X) by considering the case A=X. Since \mathrm{Hom}(X,X) has a distinguished element, the identity map X\to X, we get a corresponding distinguished element x_0 of F(X).

You can check that this is where the distinguished data on our universal objects came from. This amounts to the following trivial facts:

  • The canonical projections \prod_{n=1}^\infty X_n\to X_1, X_2,\dots are the ones that zip together into the identity map \prod_{i=1}^n X_n\to \prod_{i=1}^n X_n.
  • The element x\in \mathbb{R}[x] is the image of x under the identity map \mathbb{R}[x]\to \mathbb{R}[x].
  • The inclusion of \mathrm{ker}(g) into B is the composite of the inclusion \mathrm{ker}(g)\to B with the identity map on \mathrm{ker}(g)\to \mathrm{ker}(g).

Yes, the canonical piece of data on the universal object is usually not very interesting in itself. But in fact, every piece of data on every object can be obtained from it!

  1. Every piece of data a\in F(A) is either the pullback or pushforward (according to whether F is contravariant or covariant) of the universal piece of data x_0\in F(X), and in a unique way.

Indeed, let’s say F is contravariant, and we have some piece of data a\in F(A) for some object A. Then a corresponds to some map f: A\to X by the universal property. What happens when we pull back x_0\in F(X) along f: A\to X? In terms of hom-sets, the pullback map F(X)\to F(A) is just the precomposition map \underline\ \circ f: \mathrm{Hom}(X,X)\to \mathrm{Hom}(A,X), so the image of x_0 in F(A) is the element corresponding to the composite \mathrm{id}_X\circ f = f, namely the element a.

In the case of products, this tells you that every tuple of maps A\to X_1,X_2,\dots arises as the universal tuple \prod_{n=1}^\infty X_n\to X_1, X_2,\dots precomposed with a unique map A\to\prod_{n=1}^\infty X_n. In the case of polynomial rings, it tells you that for any element a of an \mathbb{R}-algebra A, there’s a unique \mathbb{R}-algebra homomorphism \mathbb{R}[x]\to A sending x to a.

This is how mathematicians tend to think about universal objects: they are universally equipped with some data, in that every other object with that data gets its data via pullback or pushforward from an appropriate map to the universal object:

  • The topological product \prod_{n=1}^\infty X_n is the universal topological space equipped with continuous maps to each X_n.
  • The kernel of g:B\to C is the universal group equipped with a group homomorphism to B whose composite with g is trivial.
  • The \mathbb{R}-algebra \mathbb{R}[x] is the universal \mathbb{R}-algebra equipped with a distinguished element.

Question: You keep saying “the” universal object with such-and-such property. How do I know there’s only one? Might the box topology on \prod_{n=1}^\infty X_n have the same universal property as the product topology?

Answer: One of the strengths of universal objects is that, if they exist, they are unique up to unique isomorphism. So in particular, whenever the box topology on \prod_{n=1}^\infty X_n is different from the product topology, it won’t have the property that a function to it is continuous if and only if each component is continuous.

Here’s the proof that any two objects X and Y with the same universal property are isomorphic. Say maps to both X and Y parameterize the contravariant functor F—in particular, we are assuming that F already comes with a notion of pulling back pieces of data. Then the distinguished element y_0\in F(Y) gives a map f:Y\to X such that y_0 is the pullback of x_0\in F(X). Similarly, x_0 gives a map g:X\to Y making x_0 the pullback of y_0\in F(Y). Then the composite f\circ g:X\to X is a map pulling back x_0 to itself; it must therefore equal the identity map \mathrm{id}_X:X\to X which has that same property. Similarly, g\circ f:Y\to Y must equal \mathrm{id}_Y. Therefore f and g are inverses of each other and X\cong Y.

So any two objects with the same universal property are isomorphic; moreover, that isomorphism is unique if we require it to pull back or push forward the universal pieces of data to each other. (For all I know, even if the box and product topologies on a product \prod_{n=1}^\infty X_n are different, the resulting topological spaces might be abstractly isomorphic. But there won’t be an isomorphism between them that preserves each projection \prod_{n=1}^\infty X_n\to X_1,X_2,\dots, because the only function that does is the identity.)

Question: So there’s always at most one object with a given universal property, if we count isomorphic objects as “the same.” Is there always at least one universal object?

Answer: No, not every universal property is manifested by some universal object. An important case is that of fields: There is no universal field extension of \mathbb{R} equipped with a distinguished element analogous to \mathbb{R}[x] for \mathbb{R}-algebras (which don’t need to be fields). The obvious guess would be the fraction field \mathbb{R}(x), but the image of x under a field homomorphism \mathbb{R}(x)\to K can never hit zero, since 1/x also has to go somewhere. The same holds for every nonzero polynomial in x, so x can never be sent to an algebraic element over \mathbb{R}. However, that’s the only problem, so \mathbb{R}(x) is actually the universal field extension of \mathbb{R} equipped with a distinguished transcendental element!

We can identify other field extensions as having similar universal properties: \mathbb{C} is the universal field extension of \mathbb{R} equipped with a square root of -1, for example, since for any choice of field homomorphism \mathbb{R}\to K and choice of element a\in K whose square is -1, there is a unique field homomorphism \mathbb{C}\to K sending a to -1. In general, for any irreducible polynomial f(x) over a field K, the field extension K[x]/(f(x)) is the universal field extension of K equipped with a root of f.

Question: Does every mathematical object have a universal property?

Answer: Sometimes people do wonder whether a certain object they’ve discovered can be characterized in terms of a universal property. But this always means “in terms of a nice universal property,” because every object X has a kind of self-referential universal property: we could just define the data on an object A to be a map from A to X. Then X is universal with that property, and the distinguished piece of data X has is the identity map \mathrm{id}_X:X\to X. Not very interesting!


Has this post helped you, or left you with further questions? Let me know with a comment!

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